What is the integral from 7 to 8 of x/(x^2-3) ?

The integral from 7 to 8 of x/(x^2-3) is 1/2 (ln(61)-ln(46))

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integral from 7 to 8 of x/(x^2-3)

\int_{7}^{8} \frac{x}{x ^{2} - 3} d x

\frac{1}{2} (ln (61) - ln (46))

Decimal Notation

0.14111 \dots

Solution steps

\int_{7}^{8} \frac{x}{x ^{2} - 3} d x

Apply u-substitution

: \int_{46}^{61} \frac{1}{2 u} d u

= \int_{46}^{61} \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{46}^{61} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{46}^{61}

Compute the boundaries:

ln (61) - ln (46)

= \frac{1}{2} (ln (61) - ln (46))

x \to 7 + lim (3 x + 5)

x

\frac{d y}{d t} = 2 y - 3 e^{- t}

f (x) = x^{3} - 2 x + 2, at x = 2

x \to 0 - lim (- 2 x^{2} + 7 x)

What is the integral from 7 to 8 of x/(x^2-3) ?
The integral from 7 to 8 of x/(x^2-3) is 1/2 (ln(61)-ln(46))